Question

Решите неравенство:3tg(1/6x+π/6)<√3​

Answer 1

Ответ:

Решить  неравенство .

$\displaystyle\bf 3\, tg\Big(\frac{1}{6}\, x+\frac{\pi }{6}\Big) < \sqrt3\ \ \ \Rightarrow \ \ \ \ tg\Big(\frac{1}{6}\, x+\frac{\pi }{6}\Big) < \frac{\sqrt3}{3}\\\\\\-\frac{\pi }{2}+\pi n < \Big(\frac{1}{6}\, x+\frac{\pi }{6}\Big) < \frac{\pi }{6}+\pi n\ \ ,\ \ n\in Z\\\\\\-\frac{2\pi }{3}+\pi n < \frac{1}{6}\, x < \pi n\ \ ,\ \ n\in Z\\\\\\-4\pi +6\pi n < x < 6\pi n\ \ ,\ \ n\in Z\\\\\\Otvet:\ x\in \Big(-4\pi +6\pi n\ ;\ 6\pi n\ \Big)\ \ ,\ \ n\in Z$